## The Game of Hangman

• "Defendant" picks a word, writes blanks for its letters, draws a gallows

----
|  |
|
|
|
|       _ _ _ _ _ _ _ _

• "Prosecution" guesses letters of the word in turn. Guessing a letter not in the word causes a body part to be added to the gallows. Guessing a letter in the word fills in all occurrences

----   z w
|  |
|
|  O
|  |
|      _ _ _ _ _ _ i _

• "Trial" is over when word is completed or def is hanged

----   z w e a o u
|  |
|
|  O
| /|\
| / \  _ _ _ _ _ _ i c


## Hangman Prosecution and Utility

• What letter should pros guess first?

• Let's assume that def chose random 8-letter word from known dictionary

• Let's further assume that a found letter has a uniform value of 1

• Then find

argmax [l in 'a'..'z'] (l in word) pr(word|dict)

• This is an easy calculation: 'e' occurs in 2/3 of the 8-letter words (see bestguess.py)

## Hangman: My Assumptions Are Weak

• Problem: because 'e' is so common, hitting it doesn't necessarily narrow things down so much

• Value function should be based on expected chance of winning after hitting or missing the letter

• Problem: Def isn't choosing uniformly — will pick "hard" words

• Need to know probability given defense strategy pr(word|ds(dict))

• So

argmax [l in 'a'..'z'] v(win with word) pr(word|ds(dict))


## Hangman: Nash Equilibrium

• Ugh. ds(dict) will depend on guessing strategy

• "Good old rock. Can always count on rock"
• Adversary games are hard. But Nash Equilibrium exists and can in principle be calculated

• Current research for me

## Information Theory

• Consider these strings of bits

0111111111111111

0101100010011011

• x is boring and easy to describe. You could predict the "next value" pretty reliably

• y is complicated

• Shannon et al: x has "less information" than y

• Information content can be viewed as a utility function. The entropy of a set is given as

u(S) = sum [i in S] pr(i) lg 1/pr(i)
= sum [i in S] - pr(i) lg pr(i)


For our sets (strings)

u(x) = - pr(0) lg pr(0) - pr(1) lg pr(1)
= - (1/16) lg (1/16) - (15/16) lg (15/16)
~= 0.337

u(y) = - pr(0) lg pr(0) - pr(1) lg pr(1)
= - 2 (8/16) lg (8/16)
= -2 × 0.5 × -1 = 1


## Hangman: Entropy-Based Prosecution

• Pros wants to get to a state where there is only one choice

• Standard trick: pick the letter that has the greatest expected chance of reducing the entropy the most

argmax [l in 'a'..'z']
(1 - sum [p in part(l, dict)] u(l))) pr(l)

• Does this still produce 'e' as the initial guess? It does

• We have taken into account the "cost of gathering information" as part of the utility: we don't want to make a guess that costs us a body part unless we get a lot of information from it

## Demonic Hangman

• Consider "Demonic Hangman": Def cheats as desired by changing the word in a way consistent with the guesses so far

• Now we almost have to use entropy to narrow Def down to only one choice quickly

Last modified: Monday, 26 October 2020, 11:37 PM